1.7 Fundamental subspaces of an operator

Definition 1 Let $X$ be a finite dimensional Hilbert Space with orthonormal basis $\{b_1,b_2,...,b_n\}$ and $A:X\to Y$ be a linear operator. The range of $A$ is the subspace

It is easy to see that $\mathcal{R}\left(A\right)$ is a subspace of $Y$ . To show the second equality above, note that if $x\in X$ then it can be written as $x={\sum }_{i=1}^n{a}_i{b}_i$ ; therefore,

where ${\psi }_i=A{b}_i$ . Before we can claim that $\left\{{\psi }_i\right\}$ is a basis for $\mathcal{R}\left(A\right)$ , we must first show its elements are linearly independent.

Lemma 1 If $X$ be $n$ -dimensional and $A:X\to Y$ , then $\mathcal{R}\left(A\right)$ has dimension less than for equal to $n$ .

If the set $\{{\psi }_1,{\psi }_2,...,{\psi }_n\}$ given above is linearly independent then it is a basis for $\mathcal{R}\left(A\right)$ , and the dimension of $\mathcal{R}\left(A\right)$ is $n$ . If they are not linearly independent we must show that $\dim \left(\mathcal{R}\right(A\left)\right)<n$ . If $\{{\psi }_1,{\psi }_2,...,{\psi }_n\}$ are linearly dependent then there exists a set of scalars $\{{\alpha }_1,{\alpha }_2,...,{\alpha }_n\}$ such that ${\sum }_{i=1}^n{\alpha }_i{\psi }_i=0$ with at leas one nonzero ${\alpha }_i$ ; we let that be ${\alpha }_1$ without loss of generality. We then have

and so $\mathrm{span}\left(\{{\psi }_1,...,{\psi }_n\}\right)=\mathrm{span}\left(\{{\psi }_2,...,{\psi }_n\}\right)$ . If the set $\{{\psi }_2,...,{\psi }_n\}$ is linearly independent, then $\dim \left(\mathcal{R}\right(A\left)\right)=n-1\le n$ . Otherwise, iterate this procedure to show that $\dim \left(\mathcal{R}\right(A\left)\right)<n-1$ .

Definition 2 The null space of a $A$ is the set of all points $x\in X$ that map to zero:

It is easy to see that $\mathcal{N}\left(A\right)$ is a subspace of X.

Lemma 2 An operator $A$ is non-singular if and only if $\mathcal{N}\left(A\right)=\left\{0\right\}$ .

We can extend the concept of rank from matrices to operators.

Definition 3 The rank of A is the dimension of $\mathcal{R}\left(A\right)=\mathrm{rank}\left(A\right)$ .

Theorem 1 Let $A:X\to Y$ and $X$ be an n-dimensional space. Then,

Let $\mathcal{N}\left(A\right)$ have dimension $m$ . Design an orthonormal basis $e_1,...,e_n$ such that $e_1,...,e_m$ are an orthonormal basis for $\mathcal{N}\left(A\right)\subseteq X$ . This implies that ${\psi }_i=A{e}_i=0$ , for $i=1,...,m$ . Therefore, $\mathcal{R}\left(A\right)=\mathrm{span}\left({\left\{{\psi }_i\right\}}_{i=1}^n\right)=\mathrm{span}\left({\left\{{\psi }_i\right\}}_{i=m+1}^n\right)$ , so $\mathrm{rank}\left(A\right)=\dim \left(\mathcal{R}\right(A\left)\right)\le n-m$ .

Now we need to show that ${\left\{{\psi }_i\right\}}_{i=m+1}^n$ are linearly independent. We use contradiction by assuming that ${\left\{{\psi }_i\right\}}_{i=m+1}^n$ are linearly dependent, which means that ${\sum }_{i=m+1}^n{b}_i{\psi }_i=0$ for some scalars $b_i$ that are not all zero, i.e.,

so we know ${\sum }_{i=m+1}^n{b}_i{e}_i\in \mathcal{N}\left(A\right)$ . We also know that $e_i\perp e_j$ , $i\ne j$ , and so $e_i\perp \mathcal{N}\left(A\right)$ for $i=m+1,...,n$ . This implies in turn that ${\sum }_{i=m+1}^n{b}_i{e}_i\perp \mathcal{N}\left(A\right)$ for all choices of $\left\{b_i\right\}$ . Because ${\sum }_{i=m+1}^n{b}_i{e}_i\in \mathcal{N}\left(A\right)$ , the only possibility is that ${\sum }_{i=m+1}^n{b}_i{e}_i=0$ . However, the orthonormal basis vectors $e_i$ for $i=m+1,...,n$ are linear independent, and so we must have that $b_i=0$ . This is a contradiction with our original assumption, implying that the vectors ${\left\{{\psi }_i\right\}}_{i=m+1}^n$ are linearly independent. Therefore, this set of vectors is a basis for $R\left(A\right)$ and $\mathrm{rank}\left(A\right)=\dim \left(\mathrm{R}\right(A\left)\right)=n-m$ . This in turn implies that $\mathrm{rank}\left(A\right)+\dim \left(\mathcal{N}\right(A\left)\right)=n.$